# Given the plane curve, find tangent vector

1. Feb 25, 2012

### mdawg467

1. The problem statement, all variables and given/known data
Consider the plane curve $$\overrightarrow{r(t)}=e^tcost(t)\hat{i}+e^tsin(t) \hat{j}$$
Find the following when t= ∏/2
$$Part A: \hat{T}(t)$$
$$Part B: \hat{B}(t)$$
$$Part C: \hat{N}(t)$$
2. Relevant equations

$$\hat{N}(t)=\frac{\hat{T}(t)}{||\hat{T}(t)||}$$
$$\hat{T}(t)=\frac{\overrightarrow{r'(t)}}{|| \overrightarrow{r'(t)}||}$$

$$\hat{B(t)}=\frac{\overrightarrow{r'(t)\times r''(t) }}{||\overrightarrow{r'(t)\times r''(t)}||}$$

3. The attempt at a solution
Part A
$$\overrightarrow{r(t)}=e^tcost(t)\hat{i}+e^tsin(t) \hat{j}$$

$$\overrightarrow{r'(t)}=e^t[(cos(t)-sin(t))\hat{i} \:+\:(sin(t)+cos(t))\hat{j}]$$

$$\overrightarrow{r'(\frac{\pi }{2})}=e^\frac{\pi }{2}[(cos(\frac{\pi }{2})-sin(\frac{\pi }{2}))\hat{i} \:+\:(sin(\frac{\pi }{2})+cos(\frac{\pi }{2}))\hat{j}]$$

$$\overrightarrow{r'(\frac{\pi }{2})}=-e^\frac{\pi }{2}\hat{i}\;+\;e^\frac{\pi }{2}\hat{j}$$

$$\hat{T}(t) =\frac{-e^\frac{\pi }{2}\hat{i}\;+\;e^\frac{\pi }{2}\hat{j}}{ \sqrt{(-e^\frac{\pi }{2})^2\;+\;(e^\frac{\pi }{2})^2} }$$

Based off of Part A, plugging the numbers into Part B and C generate:
$$\hat{B(t)}=\frac{\overrightarrow{r'(t)\times r''(t) }}{||\overrightarrow{r'(t)\times r''(t)}||}=0$$
$$\hat{N}(t)=\frac{\hat{T}(t)}{||\hat{T}(t)||}=0$$

Not sure if I solved this correctly.

Any help would be great. Thank you.

2. Feb 25, 2012

### Dick

You got T(t) right. That's about all you've really shown. Your formula for N(t) isn't even right. It's supposed to have a derivative in it.